Question

Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.

Answer 1

The volume percent of graphite is 91.906 per cent.

Step-by-step explanation:

The volume percent of graphite (
`\% V_(Gr)`) is determined by the following expression:

`\%V_(Gr) = (V_(Gr))/(V_(Gr)+V_(Fe)) * 100\,\%`

`\%V_(Gr) = (1)/(1+(V_(Gr))/(V_(Fe)) )* 100\,\%`

Where:

`V_(Gr)` - Volume occupied by the graphite phase, measured in cubic centimeters.

`V_(Fe)` - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

`V_(Gr) = (m_(Gr))/(\rho_(Gr))`

`V_(Fe) = (m_(Fe))/(\rho_(Fe))`

Where:

`m_(Gr)`,
`m_(Fe)` - Masses of the graphite and ferrite phases, measured in grams.

`\rho_(Gr)`,
`\rho_(Fe)` - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

Let substitute each volume in the definition of the volume percent of graphite:

`\%V_(Gr) = (1)/(1 +((m_(Gr))/(\rho_(Gr)) )/((m_(Fe))/(\rho_(Fe)) ) ) * 100\,\%`

`\%V_(Gr) = (1)/(1+\left((m_(Gr))/(m_(Fe)) \right)\cdot \left((\rho_(Fe))/(\rho_(Gr)) \right))* 100\,\%`

Let suppose that 100 grams of cast iron are available, masses of each phase are now determined:

`m_(Gr) = (2.5)/(100)* (100\,g)`

`m_(Gr) = 2.5\,g`

`m_(Fe) = 100\,g - 2.5\,g`

`m_(Fe) = 97.5\,g`

If
`m_(Gr) = 2.5\,g`,
`m_(Fe) = 97.5\,g`,
`\rho_(Fe) = 7.9\,(g)/(cm^(3))` and
`\rho_(Gr) = 2.3\,(g)/(cm^(3))`, the volume percent of graphite is:

`\%V_(Gr) = (1)/(1+\left((2.5\,gr)/(97.5\,gr) \right)\cdot \left((7.9\,(g)/(cm^(3)) )/(2.3\,(g)/(cm^(3)) ) \right)) * 100\,\%`

`\% V_(Gr) = 91.906\,\%`

The volume percent of graphite is 91.906 per cent.